On Thu, Feb 6, 2025 at 4:48 AM Alan Grayson <agrayson2...@gmail.com> wrote:
> > > On Monday, February 3, 2025 at 6:44:29 AM UTC-7 Jesse Mazer wrote: > > On Mon, Feb 3, 2025 at 3:16 AM Alan Grayson <agrays...@gmail.com> wrote: > > > > On Sunday, February 2, 2025 at 11:37:48 PM UTC-7 Jesse Mazer wrote: > > On Mon, Feb 3, 2025 at 12:09 AM Alan Grayson <agrays...@gmail.com> wrote: > > > > On Sunday, February 2, 2025 at 9:39:04 PM UTC-7 Jesse Mazer wrote: > > On Sun, Feb 2, 2025 at 11:04 PM Alan Grayson <agrays...@gmail.com> wrote: > > *In mathematics, functions have domains and ranges, standard terminology. > A function maps domain sets to range sets. The image of a function is the > set containing its range. Again, standard mathematical terminology. The > contraction formula, derived for the LT, is a function. With me so far? AG* > > > In math the domain and range of a function have no physical > interpretation, they are just sets of mathematical objects with no comment > about what they "mean". For example, in purely mathematical terms, the > length contraction formula l = L*sqrt(1 - v^2/c^2) has as its domain values > of the variables L and v where L can be any member of the set of real > numbers > 0, and v is any member of the set of real numbers larger than or > equal to 0 but smaller than the value of c (in whatever units you're using > like 299792458 meter/second), and the range is the value of l which can > likewise be any member of the set of real numbers > 0. > > If you are speaking more metaphorically, basically just saying that any > equation like length contraction can be thought of as a sort of machine > that takes two values (a proper length L and a speed v) as input/"domain" > and spits out another value (a length l in whatever frame measured the > object to be moving at v) as output/"range", then I'm fine with that. > > > *Now about the substance. If coordinate frame O2 is the domain of the > formula function, the x values are elemments in its domain, and the x' > values are elements in its range.* > > > If you are talking about the LT function when applied to a problem where > we start with coordinates in O2 as input and get coordinates in O1 as > output, sure. If you're talking about the length contraction formula, no, > the only inputs to that are a proper length and a velocity in a single > frame, and the output is the length in that same frame. > > > *Here's where we disagree. ISTM, there's a convention, that the image > frame of the formula, is moving wrt the frame applying the contraction > formula,* > > > The length contraction formula is not exactly translating between > different frames at all, at least not in the sense that the input > exclusively consists of variables from one frame and the output a variable > from a different frame, the way the LT does. In the length contraction > formula, both the speed v which is used as input and the contracted length > L' which is given as output are measured in the SAME frame, the frame of > the observer who sees the object moving at speed v relative to their own > frame. The proper length L, which is also used as input, can be thought of > as length in a different frame, namely the object's own rest frame. > > Since there is a symmetry of motion where the speed of B relative to A is > the same as speed A relative to B, I suppose the v in the length > contraction formula *could* instead be defined as the speed of the observer > as measured in the object's rest frame, rather than defining it as the > speed of the object in the observer's frame as textbooks normally do. If > you think of it in that alternate way where v is the speed of the observer > in the object's frame, then you could consider both input variables L and v > to be measured in one frame (the object's rest frame) and the output > variable L' in another (the observer's frame). Even though v is not > normally described this way, it'd make no difference mathematically if you > did. But if we do think of it this way, I presume you'd then say the > object's rest frame is the frame "applying the contraction formula", and > the observer's frame is the "image frame"? > > > > * and L' is the contracted length in the frame in relative motion. If you > claim the contraction occurs in the same frame from which the formula is > applied, then won't we get no contraction? AG * > > > See above, the normal way of describing the length contraction formula > involves a mix of measurements in two different frames as input so there is > no single 'frame from which the formula is applied', but if you want, you > do have the option to think of the meaning of v in a different way such > that both L and v are defined in terms of measurements in a single input > frame (the rest frame of the object), in which case the contracted length > L' would be in a different output frame (the rest frame of the observer). > > > > > Also note that there is nothing in the LT formula that restricts which > frame you take as input and which you get as output. You can just as easily > start with the coordinates in O1 as input and use the formula to find the > coordinates in O2 as output. I just said this in the post above and even > offered to give you a numerical example of how this would work. > > > *Of course, but why do that? AG * > > > Because you asked how the LT could be used to predict a contracted > length--in order to do that, the output of the LT formula has to be defined > in a frame where the object has non-zero velocity. > > > > > * The contraction formula is a mapping or correspondence from coordinate > sets O2 to coordinate sets O1, the moving frame in relative motion wrt O2.* > > > The contraction formula doesn't do that, the LT *can* be used to do that > if you start with coordinates in O2 as input and then get coordinates in O1 > as output, but as I say above it can just as easily map coordinates in O1 > taken as input to coordinates in O2 given as output. > > > * That is, the contraction formula maps O2, the frame with no rod, to O1, > the frame with the rod.* > > > I asked you several times in the last comment (and in a number before > that) to clarify if by "no rod" you just mean the rod is not *at rest* in > O2, or if you mean that O2 literally doesn't "see" the rod to assign it > coordinates, or something else. > > > *There is one rod is at rest in O1, and its frame in moving relatively wrt > O2. AG* > > > But you agree that the observer O2 can measure and assign coordinates to > the rod in O2's own rest frame? > > > > > * You say, and I now agree, that there's no contraction of the one and > only rod in O1.* > > > Yes. > > > * So what happened to contraction?* > > > The rod is contracted in other frames like O2, and as I said you're free > to use the LT to start with the coordinates in O1 used as input and then > use the LT formula to get the coordinates in O2 as output (once you know > the coordinates of the front and back of the rod in O2 it's easy to get the > length of the rod in O2 from that). And if you use the length contraction > formula rather than the LT it's an even simpler matter to derive the rod's > contraction in O2, you just need the rod's proper length L as well as its > velocity v in O2, you enter that L and v into the length contraction > formula as input and get the contracted length l in O2 as output. > > > *When we map from O2 to O1, you agreed, no contraction, so if we map from > O1 to O2, won't there also be no contraction?* > > > No, since the rod is at rest in O1 and moving in O2, if you map from O1 to > O2 using the LT you get a contracted length for the output. As I said I > could give you a numerical example showing this if you want. > > > *If rod is at rest in 01 and in relative motion wrt O2 (because O1 is the > moving frame) and we map from 02 to 01 we get no contraction, but if we map > in opposite direction, from O1 to O2, we get contraction? Could you explain > how you reach these conclusions? AG * > > > The quickest way to get this conclusion is just to know that the > predictions of the LT about length will always match those of the length > contraction equation which was derived from it, and if the rod has some > known proper length like L=10 and some nonzero velocity like v = 0.6c > relative to O2, then in O2 frame its length is 10*sqrt(1 - 0.6^2) = 8, but > since it has a velocity of v = 0 relative to O1, then in the O1 frame it > must have length 10*sqrt(1 - 0^2) = 10. The LT will agree with this so if > you use the equations with O2 as input and O1 as output, you'll have a > length of 8 as input and a length of 10 as output (no contraction in output > frame). But if you use the LT equations with O1 as input and O2 as output, > you'll have a length of 10 as input and a length of 8 as output > (contraction in the output frame). > > If you don't want to just trust the principle of "the LT agrees with the > length contraction equation", you can verify this by direct calculation > using the LT equations. Let's call O2 the unprimed frame and O1 the primed > frame. In the O2 frame we have the following equations for the position as > a function of time of each end of the rod (i.e. the worldline of each end): > > Back of the rod: x = 0.6c*t > Front of the rod: x = 8 + 0.6c*t > > These equations tell you that at t=0 the back of the rod is at x=0 and the > front of the rod is at x=8, so the rod has a length of 8 in the O2 frame, > and it's moving at 0.6c. > > Then in the primed O1 frame we have the following equations: > > Back of the rod: x' = 0 > Front of the rod: x' = 10 > > These equations tell you that at any choice of time coordinate t' the back > is always located at x'=0 and the front is always located at x'=10, i.e. > the rod is at rest in this frame and has a length of 10 (its proper length). > > One way to use the LT is to directly plug the full equations of motion in > one frame into the LT equations as input, and after a little algebra get > out the equations of motion in the other frame as output. That's what I did > in that post at > https://groups.google.com/g/everything-list/c/ykkIYDL3mTg/m/giZVF9PpDQAJ > going from the equations of motion in the frame where the rod was moving > (here the O2 frame) to the equations of motion in the frame where the rod > was at rest (O1). This does involve a little algebra though. A simpler way > of just checking that the LT map between those equations of motion is just > to pick the coordinates of some individual points that are along a given > end's worldline in the coordinates of one frame, and see that they always > map to coordinates of individual points that are along the same end's > worldline in the coordinates of the other frame. > > For example in the O2 frame, the points (x=0, t=0) and (x=3, t=5) and > (x=6, t=10) and (x=9, t=15) would all lie along the worldline of the BACK > of the rod given by x = 0.6c*t in this frame (I'm assuming we're using > units like light-seconds and seconds where c=1). If you take any of those > points as input and use the x-->x' LT to find the corresponding coordinates > of the point in the primed O1 frame as output, you will get (x'=0, t'=0) > and (x'=0, t'=4) and (x'=0, t'=8) and (x'=0, t'=12) [note that the x-->x' > equation in this case is x' = 1.25*(x - 0.6c*t) and the t-->t' equation is > t' = 1.25*(t - 0.6*x)]. You can see that all of these points do lie along > the line x'=0, the equation for the BACK of the rod in the O1 frame. > > Similarly, in the O2 frame, the points (x=8, t=0) and (x=11, t=5) and > (x=14, t=10) and (x=17, t=15) all like along the worldline of the FRONT of > the rod given by x = 8 + 0.6c*t in this frame. If you take any of those > points as input and use the x-->x' LT to find the corresponding coordinates > in the primed O1 frame as output, you will get (x'=10, t'=-6) and (x'=10, > t'=-2) and (x'=10, t'=2) and (x'=10, t'=6). You can see all these points > lie along the line x'=10, the equation for the FRONT of the rod in the O1 > frame. > > Then if you want to go in reverse, starting with O1 coordinates as input > and getting O2 coordinates as output, you can use the LT equations for > x'-->x which in this example is x = 1.25*(x' + 0.6c*t'), and the one for > t'-->t which is t = 1.25*(t' + 0.6*x'). You can verify that the reverse > mapping here works too, for example if you take (x'=10, t'=2) as input (a > point I listed above as being on the worldline of the front of the rod) you > get back (x=14, t=10) as output. So you can verify this way that if you are > given the equations x' = 0 and x' = 10 for the rod in the O1 frame > (corresponding to a rod at rest with length 10 in the O1 frame), if you > pick any point along those lines and map to the O2 frame with the LT, as > output you always get points along x = 0.6c*t and x = 8 + 0.6c*t > (corresponding to a rod of length of length 8 and velocity 0.6c in the O2 > frame). So, here the input coordinates represent a rod with its proper > length of 10, and the output coordinates represent a rod with a contracted > length of 8. > > Jesse > > > Let's start from the beginning to make sure we're on the same page. Using > the contraction formula L' = L * sqrt ( 1 - (v/c)^2), where L is the rest > length of rod in frame f1, moving at velocity v wrt frame f2, and L' is the > contracted length of the rod as calculated from the pov of f2. Do you agree > with my interpretations of these variables? Do you agree that the measured > length of rod in f1 is never L', but always its rest length L? These are > Yes or No questions. TY, AG > As long as we are considering a case where f2 is different from f1 (which is normally the only situation where anyone would bother to use the formula), I agree with everything you just wrote. But the one caveat I'd add is that one is technically free to consider the special case where f2 = f1, i.e. you are imagining that the frame f2 of the observer who is using the equation is the same as f1, the rest frame of the rod, so in this case v=0 and L' = L. In my first paragraph above (the one starting with 'The quickest way to get this conclusion') I had L=10, and I considered both the case of an observer O2 moving at 0.6c relative to the rod who used L=10 and v=0.6c as input to the formula to output a length L' = 8 in his frame, and also the case of observer O1 who was at rest relative to the rod and used L=10 and v=0 as input to the formula to output a length of L' = 10 in his frame. Of course the latter is a trivial case and it's easier to just remember that rest length/proper length is what will be measured in the frame where the rod is at rest, but I included it to illustrate the point that either observer can apply the formula to get the correct length in their frame as the output. Jesse > > > > > > * And if we map from O2 to O2, using the rod's coordinates in O2 as input, > won't there also be no contraction? AG * > > > In this case there is no change in the length from input to output, but > that length is still contracted relative to the object's proper length > (this calculation going from O2 to O2 wouldn't have the proper length > appearing on either side of the equation though, you'd need to do a > different calculation to find it). > > Jesse > > > > * So, AFAICT, contrary to what relativity claims, contraction doesn't > exist! Note also what happens to the Parking Pardox. No contraction of any > object in any frame. Paradox solved! Are we having fun yet? AG* > > > On Sunday, February 2, 2025 at 7:52:55 PM UTC-7 Jesse Mazer wrote: > > On Sun, Feb 2, 2025 at 8:17 PM Alan Grayson <agrays...@gmail.com> wrote: > > On Sunday, February 2, 2025 at 4:31:48 PM UTC-7 Jesse Mazer wrote: > > On Sun, Feb 2, 2025 at 5:08 PM Alan Grayson <agrays...@gmail.com> wrote: > > On Sunday, February 2, 2025 at 2:43:09 PM UTC-7 Jesse Mazer wrote: > > On Sun, Feb 2, 2025 at 12:44 PM Alan Grayson <agrays...@gmail.com> wrote: > > I will study your post and respond later. For now, let me say that the GPS > situation is irrelevant. It just shows that time dilation is real. Nothing > to do with length contraction. Also, after reading some of your earlier > comments, I agree that in the frame containing the rod, its length is not > contracted. This is the rest frame with the rod at the origin. The frame > from which the LT is applied has an observer at the origin, but no rod, and > is in relative motion compared to the frame with the rod. I hope you have > no objections to this comment. If you have any objections, please let me > know. AG > > > Mostly sounds fine but the only thing I'd want to double check is that > when you say "The frame from which the LT is applied has an observer at the > origin, but no rod", do you just mean that the rod is not at rest in this > observer's frame? The rod is still measurable and can be assigned > coordinates with changing position as a function of time in this observer's > frame (the observer you called O2 in your earlier post), agreed? > > > I want two frames with the rod in one, which I thought was your initial > model. The rod is situated and fixed at the origin, and there is no rod in > the frame using the LT; > > > But as I asked you repeatedly, when you say no rod "in" the O2 frame do > you just mean there is no rod that's *at rest relative to* the O2 frame, or > are you somehow denying that any given physical object like a rod is > assigned coordinates by *all* frames including the O2 frame in which the > rod is moving? > > > You still haven't answered this question, and it seems like it might be > important given some of your other phrases below... > > > > > or if you prefer we can model the situation with a rod in each frame, at > rest, both at origin, and their rest lengths are unimportant. > > > No need for two rods, provided you agree above that the O2 frame still > assigns coordinates to the rod even though the rod is not at rest in that > frame. > > > *OK. One rod, and frame with rod is given coordinates in both frames. For > me, x ---> x' means a LT from frame with no rod, * > > > Does "frame with no rod" just mean "frame with no rod at rest in it", or > are you somehow claiming there's a frame that doesn't "see" the rod at all > in terms of being able to measure it and assign coordinates to it? > > > > *to frame with rod, and from this there's the implication of contraction > in x' frame, with rod. Do you agree or not? AG* > > > > Drawing on the GPS situation, from any rod/frame applying the LT, the > formula IMO predicts the measured length in the frame it is observing, > > > GPS is distinct because the clocks don't just tick at their natural rate > > > *They tick naturally and are then reset to presumably synchronize them > with orbiting clock. AG* > > but are artificially adjusted, as I said. If the rod is at rest in O1 and > moving relative to O2, can we assume we are initially given the coordinates > of the rod as measured in O2, then then O2 frame is the one "applying" the > LT to predict the coordinates in the frame O1, so that O1 would be "the > frame it is observing" in your statement above? > > > *Yes, except we don't have to assume the moving rod has coordinates in O2. > AG * > > > Do you just mean it doesn't have *fixed* coordinates in O2, or do you mean > it isn't assigned coordinates at all in O2? If the latter, are you > imagining it's somehow invisible to the O2 observer? If so that's not how > things work in relativity, the rod is just an ordinary physical object, of > course the O2 observer is going to be able to measure it as it passes by > his own system of rulers and clocks, and say things like "when the clock > attached to the 3-light-second mark on my ruler showed a time of 5 seconds, > the back of the rod was passing right next to it (as seen in a photo taken > at that location at that moment, for example), therefore the worldline of > the back of the rod passes through the coordinates x=3 light seconds, t=5 > seconds in my coordinate system" > > > > > > similar to the Earth bound clocks in GPS which predict the time delays in > the orbiting clocks. For this reason, in the contraction case, the > frame/observer applying the LT, doesn't predict the contraction in > observer's own frame (which doesn't exist if there's one rod in the model), > but in the frame with the rod. > > > No, as my numerical example shows, if we start with the coordinates of the > rod in O2 and use the LT to predict its coordinates in O1, we get a > prediction of NO contraction of the rod in O1; the prediction will be that > the rod has its "proper length" in the O1 frame. > > > However, and this is where I get my prediction which you object to; in > this frame, the frame with the rod, the only prediction possible is zero > contraction. > > > If you are talking about the type of calculation I describe above, I > *agree* the prediction would be zero contraction in the O1 frame, which > matches the fact that no contraction is MEASURED in the O1 frame. > > > *Yes, this is what I've been saying. AG* > > > It was you who claimed that there was some prediction using the LT that > would conflict with the fact we both agree on that no contraction is > measured in the frame where the rod is at rest. > > > *I changed my pov when reading one of your previous posts. But since > there's no contraction measured in frame where the rod exists* > > > Are you saying the rod literally does not "exist" in other frames in the > sense of not being measured at all, or are you just saying the rod is not > at rest in other frames? If you're somehow saying the rod is not assigned > coordinates at all in the O2 frame, that doesn't make sense, see above. > > > > * and is at rest (even though the frame is in relative motion), the LT has > no other possible predictions, so it seems that length contraction never > occurs!* > > > Sure length contraction occurs, in the example it occurs for the O2 > observer who sees the rod in motion. If the rod has a proper length of 10 > light-seconds, and the O2 observer says the BACK of the rod passed by x=3 > and t=5 in his coordinate system, and the FRONT of the rod passed by x=11 > and t=5 in his system, then the distance between the front and back at the > single moment t=5 in this frame must be 11 - 3 = 8, so that's what the O2 > judges the rod's length to be, a length which is shorter than the rod's > proper length measured by the O1 observer. > > If you want to start from coordinates in one frame and then use the LT to > predict coordinates in another, and you're asking about how this could ever > lead to a prediction of contraction, one option would be to *start* from > the coordinates in O1's frame where the rod is at rest (unlike in my > numerical example where I started with the rod's coordinates in the frame > where the rod was moving), then apply the LT to *predict* the coordinates > in O2's frame where the rod is moving. If we did that instead of the > opposite, in that case we *would* get a prediction of a contracted length > in the predicted frame O2. I could give a different numerical example > illustrating this, if you would actually be interested in reading through > it. > > > > * This is where the rubber hits the road in our disagreement about what > the LT predicts, and what is measured. If contraction is never measured > because it never occurs, the "predictions" of the LT are worthless to the > point of not existing. I hope you're not going to tell me now, that x' in x > --> x' refers to the frame applying the LT. AG* > > > > Hence, the LT doesn't do as you claim, and it doesn't function like the > GPS situation. Moreover, I recall you used spacetime diagrams to show > length *expansion *in your frame at relative speed, but never before have > I heard or read of such a claim, which raises the proverbial red flag. AG > > > As I stated repeatedly, by "expansion" I just meant the length would be > predicted as larger in the second frame (O1 above) compared to the first > frame where the rod was moving (O2 above), > > > *How could that be if the result of the LT is x', not x?* > > > In my numerical example I treated the primed frame as the one where the > rod is at rest, i.e. O1 in your terminology. So, going from x-->x' takes > you from O2 to O1. > > Also note however that the complete set of LT equations include both > x-->x' and x'-->x, so you're free to go in either direction. > > > > * The rod is not contracted in O1 even though its moving relative to O2, > and the length of the rod is not in the image of anything in O2?* > > > If you want to do the LT from O2 to O1 you have to start with the > coordinates of the rod in O2, I don't know if that's what you mean by "the > image of anything" or if you just mean that we don't obtain the O2 > coordinates as a result of a LT. > > > > * Remember; the LT prediction in O1 is no contraction. I think this is > exactly where we disagree about what the formula predicts.* > > > What do you think we are disagreeing about in terms of what the formula > predicts? I've said that if you start with the coordinates in O2 and use > the LT to predict O1, the prediction is no contraction in O1. > > Jesse > > > > * Maybe in your reply we can finally resolve this issue. AG* > > I wasn't talking about expansion relative to the rod's proper length. The > prediction will be that the rod has its proper length in frame O1, i.e. "no > contraction" relative to the proper length. > > Jesse > > -- > You received this message because you are subscribed to the Google Groups > "Everything List" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to everything-list+unsubscr...@googlegroups.com. > To view this discussion visit > https://groups.google.com/d/msgid/everything-list/b362c4c1-4de6-45d7-804e-34f21529c823n%40googlegroups.com > <https://groups.google.com/d/msgid/everything-list/b362c4c1-4de6-45d7-804e-34f21529c823n%40googlegroups.com?utm_medium=email&utm_source=footer> > . > -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. 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